My first intuition was to take two languages $L_1$ and $L_2$ (symbol $d$ at the end is to fulfill prefix property):
$$L_1 = \{ a^i b^i c^j d : i,j \ge 0 \} \mathrm{\ \ and\ \ } L_2 = \{ a^i b^j c^j d : i,j \ge 0 \}$$
So their union would be: $ L = L_1 \cup L_2 = \{ a^i b^i c^k d : i=j \lor j=k \} $
Then to prove $L$ not even being DCFL (e.g. in similar way as Example 10.1 from "Introduction to Automata Theory, Languages, and Computation" by Hopcroft & Ullman).
Big inconvenience would be the need to include in the proof also the proofs of all other properties (assuming this class is - just as "regular" DCFL - closed under complement etc.)
But I was told that precisely because the question is about DCFL being accepted by empty stack, the proof is actually trivial and could be expressed in just few sentences.